Question

7. Determine the AG at 298 K for the formation of ozone (O3) from oxygen if the partial pressure of oxygen is 0.210 atm, free radical oxygen (O) is 1.00x10 atm and the partial pressure of ozone is 5.00x10-7 atm. O2(g) + ( + 03(8) O2(g) Og AG (kJ/mole) 0 249.17 142.7 O3g)

Answer 1

The equation of gibbs free energy is given by

ΔG = ΔG° - RT ln(Qc)

We need ΔG° and Qc.

Qc = Qp x (RT)Δng

Δng = change in gaseous moles = 1-2 =-1

R = 0.0821 atm L / K mol is used as pressure is in atm

T = 298K

Given pO3 = 5 x 10^ (-7) atm

pO2 = 0.210 atm

pO = 1 x 10^(-5) atm

Qp = reaction quotient in terms of pressure

= pO3 / (pO2 x pO) = 0.238 atm-1

from above equation,

Now Qc = reaction quotient in terms of concentration = 0.238 /( 0.0821 x 298) = 9.7 x 10^(-3)

As Δng was -1 the RT factor comes in denominator.

And

ΔG° = ΔG° of product - ΔG° of reactant

= 142.7 - 249.17 = -106.47 kJ /mol

Now, as we have aquired all the missing quantities,

ΔG = ΔG° - RTlnQc

= -106.47 - [8.314 x 298 x ln( 9.7 x 10^(-3))÷ 1000]

= -106.47 + 11.48 = -94.99kJ/mol

R= 8.314 J /K mol is taken here as free energy is in joules and is divided by 1000 for conversion in kJ

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